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Chapter 1: Problem 280

For the following exercises, solve the exponential equation exactly. $$ 3^{x / 14}=\frac{1}{10} $$

Short Answer

Expert verified
\(x \approx -29.3566\)

Step by step solution

01

Rewrite the Equation

The given exponential equation is \(3^{x/14} = \frac{1}{10}\).\ To solve for \(x\), we need to isolate it. We begin by rewriting the equation using logarithms. The equation becomes \(\log_{3}(\frac{1}{10}) = \frac{x}{14}\).
02

Solve for x

We have \(\frac{x}{14} = \log_{3}(\frac{1}{10})\). To solve for \(x\), multiply both sides by 14: \[x = 14 \cdot \log_{3}(\frac{1}{10})\].
03

Change of Base Formula

To compute \(\log_{3}(\frac{1}{10})\), apply the change of base formula: \(\log_{3}(a) = \frac{\log_{b}(a)}{\log_{b}(3)}\).Using base 10, it becomes \(\log_{3}(\frac{1}{10}) = \frac{\log_{10}(\frac{1}{10})}{\log_{10}(3)}\).
04

Calculate Logarithms

Compute \(\log_{10}(\frac{1}{10})\) and \(\log_{10}(3)\): - \(\log_{10}(\frac{1}{10}) = -1\) because 10 raised to the power of -1 is \(\frac{1}{10}\).- Use a calculator for \(\log_{10}(3)\) to find its approximate value, \(\approx 0.4771\).
05

Substitute and Simplify

Substitute the values back into the equation for \(x\): \[x = 14 \cdot \frac{-1}{0.4771}\].Thus, \(x \approx 14 \times -2.0969\) = -29.3566.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Change of Base Formula
When dealing with logarithms that aren't in base 10, the change of base formula can be a lifesaver. This formula allows us to convert a logarithm of any base into a more manageable form, often using common bases like 10 or the natural base, e. The change of base formula is:
  • \( \log_{b}(a) = \frac{\log_{k}(a)}{\log_{k}(b)} \)
For our problem, we use base 10, which is common on scientific calculators, to convert \( \log_{3}(\frac{1}{10}) \).
  • This simplifies to: \( \frac{\log_{10}(\frac{1}{10})}{\log_{10}(3)} \).
By using the base 10 logs, we can easily compute both parts, making otherwise complex logarithms much simpler to handle.
Logarithms
Logarithms are the opposite operation of exponentiation, functioning as powerful tools in solving exponential equations. They are perfect for finding the unknown exponent in equations like \(3^{x/14}=\frac{1}{10}\). Using logarithms, we're able to "bring down" the exponent so it can be more easily isolated and solved.
  • The idea is \( \log_{b}(y) \, answers: \, b^{?}=y \)
In our equation, using a logarithm helps transform it so that \(x\), which is initially stuck as an exponent, can be moved down into a more familiar linear equation context. This transformation is crucial because it makes the equation easier to manipulate and solve.
Solving for x
In equations involving exponential expressions, the goal is often to find the unknown variable, typically represented as \(x\). Once we transform our original equation into a logarithmic form, we can directly solve for \(x\) by
  • first using the change of base formula,
  • computing the necessary logarithm values,
  • and then performing simple algebraic operations like multiplication or division.
In our exercise, after redefining and calculating \(\log_{3}(\frac{1}{10})\), we multiply by 14 to solve for \(x\). This approach requires careful calculation and understanding of logarithmic properties, yet it ultimately simplifies what would otherwise be a complex algebraic expression to solve manually.
Isolation of Variable
Isolating the variable in question, particularly \(x\) in our exponential equation, is a step-by-step process. This process ensures that \(x\) stands alone on one side of the equation, making it possible to determine its exact value. Here's how we can do it:
  • First, use logarithms to rearrange the equation so that the variable \(x\) is no longer an exponent.
  • Apply operations systematically, like multiplying by 14 to clear fractions tied to \(x\).
    • Each of these steps effectively "undoes" operations affecting \(x\), such as division or multiplication on each side.
    • Be mindful to accurately carry out operations on both sides of the equation. This balance maintains the equality as you solve.
    Not only does this build the calculation framework, but it also deepens understanding of how each step contributes to isolating \(x\). Once fully isolated, \(x\) can be computed with precision using a calculator or specific logarithmic values, resolving the exponential equation accurately.

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Most popular questions from this chapter

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